Heat Transfer by Conduction (Fourier's Law)
Calculate heat conduction using Q = kA(T₁-T₂)/d.
Fourier's law for steady-state heat transfer through solid materials.
The Formula
Fourier's law of heat conduction describes the rate of heat transfer through a solid material.
Heat flows from the hotter side to the cooler side.
The rate depends on the material's thermal conductivity, the area, the temperature difference, and the thickness.
Variables
| Symbol | Meaning |
|---|---|
| Q | Rate of heat transfer (Watts, W) |
| k | Thermal conductivity of the material (W/(m·K)) |
| A | Cross-sectional area through which heat flows (m²) |
| T₁ | Temperature of the hot side (°C or K) |
| T₂ | Temperature of the cold side (°C or K) |
| d | Thickness of the material (metres, m) |
Common Thermal Conductivities
| Material | k (W/(m·K)) |
|---|---|
| Copper | 385 |
| Aluminium | 205 |
| Steel | 50 |
| Glass | 1.0 |
| Brick | 0.7 |
| Wood | 0.15 |
| Fibreglass insulation | 0.04 |
Example 1
A glass window is 1.2 m × 0.8 m and 6 mm thick. The inside temperature is 22°C and outside is 5°C. How much heat is lost? (k = 1.0 W/(m·K))
A = 1.2 × 0.8 = 0.96 m²
d = 0.006 m
Q = kA(T₁ - T₂) / d
Q = 1.0 × 0.96 × (22 - 5) / 0.006
Q = 1.0 × 0.96 × 17 / 0.006
Q = 2,720 W (2.72 kW)
Example 2
A brick wall is 3 m × 2.5 m and 0.23 m thick. Inside is 20°C, outside is 0°C. Find the heat loss rate. (k = 0.7 W/(m·K))
A = 3 × 2.5 = 7.5 m²
Q = kA(T₁ - T₂) / d
Q = 0.7 × 7.5 × (20 - 0) / 0.23
Q = 0.7 × 7.5 × 20 / 0.23
Q ≈ 456.5 W
When to Use It
Use Fourier's law when you need to:
- Calculate heat loss through walls, windows, or insulation
- Design heating and cooling systems
- Select insulation materials and thicknesses
- Analyse thermal performance of building envelopes
This formula applies to steady-state conduction through flat surfaces.
For cylindrical objects (like pipes), a modified version using logarithmic terms is needed.